Question 1
each question on a new sheet. Give complete reasons and state theorem hypotheses.
Question 1
(a) Let S be all sequences with terms in {0, 1, 2}; let T ⊂ S consist of sequences eventually
equal to 2; and for fixed N , let U ⊂ S consist of sequences equal to 2 for every n ≥ N .
N
Classify S, T, U as finite, countably infinite, or uncountable, with proof. [5 marks]
N
(b) For a real sequence (a ) identify the property expressed by each statement:
n
(i) ∃a ∀ε > 0 ∃N ∀n ≥ N : |a − a| < ε;
n
(ii) ∀N ∃a ∀n ≥ N : a = a;
n
(iii) ∃M > 0 ∀n : |a | ≤ M ;
n
(iv) ∃a ∀N ∀ε > 0 ∃n ≥ N : |a − a| < ε.
n
[4 marks]
(c) Let A = {(−1)n(1 − 1/n) : n ≥ 1}. Prove that sup A = 1. [4 marks]
(d) If (a ) is unbounded, prove that it has a subsequence (a ) such that 1/a → 0. [7
n n n
k k
marks]
1
Worked solution and marking guidance
Solution.
Question 1 - Complete course-method solution
Historical official basis: 2020 Q1; 2021 Q3(b); 2022 Q1(c).
Course-method evidence: official solutions and marker comments in the uploaded 2020–
2026 papers; method families: countability_diagonal; quantifier_definition; supremum_infimum;
Cauchy_BW_subsequence.
Recognition signal
For cardinality classify infinitely many free coordinates/eventually fixed coordinates/a fixed tail; read quantifiers layer by layer; prove a supremum in two steps; for an unbounded subsequence,
make the absolute values exceed k successively.
First key step
Write T = ∪U ; for sup prove an upper bound and then ε-approximation; recursively choose indices for the unbounded subsequence.
N
(a) S is uncountable by diagonalisation (already the binary sequences form an uncountable
∪
subset). U has 3N−1 elements. Since T = U , it is countable; it is infinite.
N N N
(b) (i) convergence; (ii) constant (take N = 1); (iii) boundedness; (iv) a is a subsequential
limit, equivalently there is a subsequence tending to a.
(c) Every element has absolute value below 1, so 1 is an upper bound. Given ε > 0, choose
even 2m with 1/(2m) < ε. Then 1 − 1/(2m) ∈ A and exceeds 1 − ε.
(d) Construct n strictly increasing with |a | ≥ k. This is possible by unboundedness even
k nk
after removing finitely many terms. Then |1/a | ≤ 1/k → 0.
nk
Marking points
Cardinality classification: 5 marks; interpreting quantifiers: 4; supremum proof: 4; unbounded subsequence: 7.
Common errors
Assuming all infinite sequences over a finite alphabet form a countable set; misreading whether a depends on N in the quantifiers; only selecting values above k and ignoring unboundedness
in the negative direction.
Final self-check
Use |a | for unboundedness; U has N − 1 free coordinates.
n N
1
MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Tony study template card
Recognise the signal → write the first key step → check the theorem hypotheses → complete the course-method sequence → verify using the self-check.
Even if you cannot finish, write the definition, key theorem and first step to earn method marks.
Question 2
MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Question 2
√ √ √
(a) (i) Prove n + 1 − n ≤ 1/(2 n) for n ≥ 1. [2 marks]
∑ √
∞
(ii) Using this inequality, prove from first principles that 1/ n diverges. [6
n=1
marks]
∑
(b) Let a ≥ 0 and a → a ∈ [0, 1). Prove that ∞ an converges. [6 marks]
n n n=1 n
(c) Define {
n
−2
+ n
−1,
n even,
c =
n −n −2, n odd.
∑
De∑termine whether c
n
converges, and find with proof the radius of convergence
of c zn. [6 marks]
n
2
Worked solution and marking guidance
Question 2 - Complete course-method solution
Historical official basis: 2020 Q2; 2020 Q3(b); 2022 Q1(c).
Course-method evidence: official solutions and marker comments in the uploaded 2020–
2026 papers; method families: series_partial_sums; power_series_radius; alternating_dirichlet.
Recognition signal
Obtain a telescoping lower bound from a square-root difference; for exponential terms with a → a < 1 use eventual geometric comparison; split a power series defined by parity into two
n
subseries.
First key step
First rationalise; choose A ∈ (a, 1); separate the even and odd terms.
(a)(i) Rationalising,
√ √ 1 1
n + 1 − n = √ √ ≤ √ .
n + 1 + n 2 n
√ √ √
(a)(ii) Therefore 1/ n ≥ 2( n + 1 − n), so the partial sums satisfy
∑N
1
√
√ ≥ 2( N + 1 − 1) → ∞.
n
n=1
This also proves directly that the partial sums cannot converge to a finite number.
(b) Choose A with a < A < 1. Eventually a ≤ A, so the tail is dominated by the geometric
∑ n
series An. The finite initial segment is irrelevant.
(c) The alternating n−2 part converges absolutely. The remaining positive terms over even n
∑ ∑
are 1/(2m), so c diverges. For |z| < 1, split into even and odd subseries and use the
m≥1 n
ratio test on each; both converge absolutely. Since the series diverges at z = 1, the radius is
1.
Marking points
Square-root inequality: 2 marks; divergence from first principles: 6; eventual geometric comparison: 6; ordinary series and radius: 6.
Common errors
Reversing inequalities; merely citing the p-test instead of working from first principles; misreading an as na; incorrectly applying the alternating-series test just because signs alternate.
n n
Final self-check
Divergence at z = 1 suffices for R ≤ 1; inside the radius, absolute convergence must be proved.
2
MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Tony study template card
Recognise the signal → write the first key step → check the theorem hypotheses → complete the course-method sequence → verify using the self-check.
Even if you cannot finish, write the definition, key theorem and first step to earn method marks.
Question 3
MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Question 3
(a) For integer k ≥ 1 define f (0) = 0 and f (x) = xk sin(1/x3) for x ̸= 0. Prove continuity
k k
at 0 for every k, and differentiability at 0 for every k ≥ 2, finding f ′ (0). [5 marks]
k
(b) (i) Prove that 4x = cos x + 2 has a real solution. [2 marks]
(ii) A continuous f : [0, 10] → R satisfies f (m) = (−1)m for every integer 0 ≤ m ≤ 10.
Find the least number of zeros that must occur and justify it. [3 marks]
√
(c) Let h(x) = cos x on [−π/2, π/2].
(i) Find the second-order Taylor polynomial at 0. [3 marks]
(ii) Prove that 1 (h(0) + h(x)) ≤ h(x/2) for 0 < x < π/2. [3 marks]
2
(d) If f ∈ C4([a, b]) has five distinct zeros, prove that f (4)(c) = 0 for some c ∈ (a, b). [4
marks]
3
Worked solution and marking guidance
Question 3 - Complete course-method solution
Historical official basis: 2020 Q4; 2020 Q5; 2021 Q4; 2022 Q5.
Course-method evidence: official solutions and marker comments in the uploaded 2020–
2026 papers; method families: epsilon_delta_sequential; IVT_fixed_point; Taylor; convexity;
MVT_Rolle.
Recognition signal
For oscillation multiplied by a power at 0, use squeezing and the difference quotient; for a sign change use IVT to obtain a root; the second derivative sign gives convex/concave midpoint inequalities; for multiple zeros repeatedly apply
Rolle。
First key step
First bound | sin | ≤ 1; construct a continuous difference function; calculate the sign of h; decrease the zero count level by level.
(a) Since |f (x)| ≤ |x|k, continuity follows by squeeze. For k ≥ 2,
k
f (x) − f (0)
k k = xk−1 sin(1/x3) → 0,
x
so f ′ (0) = 0.
k
(b)(i) Let F (x) = 4x − cos x − 2. Then F (0) < 0 and F (1) > 0, so IVT applies. (ii) On each
interval (m, m + 1) the endpoint values have opposite signs, so there is at least one zero. The
ten intervals are disjoint, hence at least ten zeros; cos(πx) shows ten is attainable.
(c)(i) h(0) = 1, h′(0) = 0, h′′(0) = −1/2, so P (x) = 1 − x2/4. (ii) A direct computation gives
2
h′′(x) < 0 on the interval, so h is concave. The midpoint inequality for concave functions
gives the result.
(d) Rolle’s theorem gives at least four zeros of f ′, then three of f ′′, two of f (3), and one of
f (4).
Marking points
Continuity/differentiability: 5 marks; IVT and zeros: 5; Taylor + concavity: 6; repeated Rolle: 4.
Common errors
Applying the chain rule directly at x = 0; failing to exclude endpoint zeros when an open interval is required; reversing convexity/concavity inequalities; undercounting the zeros in repeated
Rolle applications.
Final self-check
h should be concave, so its midpoint value exceeds the average endpoint value.
3
MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Tony study template card
Recognise the signal → write the first key step → check the theorem hypotheses → complete the course-method sequence → verify using the self-check.
Even if you cannot finish, write the definition, key theorem and first step to earn method marks.
Question 4
MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Question 4
(a) Let f (x) = log x and P = (1, e1/n, . . . , e). Write L(f, P ) and U (f, P ) explicitly, compute
n n n
U − L, and prove lim n→∞ U (f, P n ) = 1. [8 marks]
∫
(b) Determine convergence of ∞ (e1/x − 1)r dx when r = 1, and prove convergence for
1
every real r > 1. [6 marks]
∑
(c) Suppose f (x) = ∞ a xn has radius of convergence R > 1 and
n=0 n
∫
1
f (x)xn dx = 0 (n ≥ 0).
−1
Prove that f ≡ 0 on [−1, 1]. [6 marks]
Worked solution and marking guidance
Question 4 - Complete course-method solution
Historical official basis: 2020 Q6; 2022 Q6.
Course-method evidence: official solutions and marker comments in the uploaded 2020–
2026 papers; method families: Darboux_sums; improper_integral; Taylor; FTC_integral_identity.
Recognition signal
For Darboux sums on a special partition use monotonicity to choose endpoints; compare improper integrals with 1/xr; for power-series orthogonality substitute the truncated polynomial
and then use uniform convergence to obtain the integral of the square.
First key step
Write x = ei/n; squeeze using ey ≥ 1 + y and MVT; let P be the Taylor truncation.
i N
(a) Since log x is increasing, with x = ei/n,
i
e1/n − 1 ∑n−1 e1/n − 1 ∑n−1
L = iei/n, U = (i + 1)ei/n.
n n
i=0 i=0
Thus
e1/n − 1 ∑n−1 e − 1
U − L = ei/n = → 0.
n n
∫ i=0
The common limit is e log x dx = 1.
1
(b) Since ey ≥ 1 + y, e1/x − 1 ≥ 1/x, so r = 1 diverges. By the Mean Value Theorem, for x ≥ 1,
e1/x − 1 = ec/x ≤ e/x. Hence for r > 1 the integrand is at most er/xr, which is integrable.
∑ ∫
(c) Let P (x) = N a xn. Then f P = 0. Since R > 1, P → f uniformly on [−1, 1], and
N n=0 n ∫ N N
f P → f 2 uniformly. Hence 1 f 2 = 0. The continuous non-negative function f 2 must vanish
N −1
identically.
Marking points
Upper and lower sums: 8 marks; improper integral: 6; power-series orthogonality: 6.
Common errors
Forgetting the Darboux interval lengths are not constant 1/n; stating an asymptotic inequality for r > 1 without proof; interchanging limit and integral without stating uniform
convergence.
Final self-check
∫
U − L should simplify to (e − 1)/n; e log x = 1.
1
4
MATH40002 Analysis 1 REVISED HISTORICAL COVERAGE
Tony study template card
Recognise the signal → write the first key step → check the theorem hypotheses → complete the course-method sequence → verify using the self-check.
Even if you cannot finish, write the definition, key theorem and first step to earn method marks.